Question

The derivatives of ${\sec }^{-1}\left[\frac{1}{(2x^2-1)}\right]$ with respect to $\sqrt{1-x^2}$ at $x=\frac{1}{2}$, is

Answer 1

Putting $x=\cos \theta$, we get
$u={\sec }^{-1}\frac{1}{2{\cos }^2\theta -1}={\sec }^{-1}\left(\sec 2\theta \right)=2\theta$ and $y=\sqrt{1-x^2}=\sin \theta$
$\therefore$ $u=2{\sin }^{-1}y\Rightarrow \frac{du}{dy}=\frac{2}{\sqrt{1-y^2}}=\frac{2}{\sqrt{x^2}}$
Thus, ${\begin{array}{c}\frac{du}{dy}\end{array}|}_{x=1/2}=4$